leetcode-598. Range Addition II

598. Range Addition II

Given an m * n matrix M initialized with all 0's and several update operations.

Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which means M[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.

You need to count and return the number of maximum integers in the matrix after performing all the operations.

Example 1:

Input: m = 3, n = 3operations = [[2,2],[3,3]]Output: 4Explanation: Initially, M = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]After performing [2,2], M = [[1, 1, 0], [1, 1, 0], [0, 0, 0]]After performing [3,3], M = [[2, 2, 1], [2, 2, 1], [1, 1, 1]]So the maximum integer in M is 2, and there are four of it in M. So return 4.

Note:

1. The range of m and n is [1,40000].
2. The range of a is [1,m], and the range of b is [1,n].
3. The range of operations size won't exceed 10,000.

题解：

给定一个m*n的二维数组，初始化为0

再给定一个二位数组（例如 M= { [2,2] , [3,3] }）,每次将M[i]行,M[j]列内的树加1，求最大元素的个数

只要遍历二维数组，每次取出行的最小值，列的最小值，这些就是加1次数最多的数，就是最大值，再返回他们的个数（最小行*最小列）即可。

class Solution {    public int maxCount(int m, int n, int[][] ops) {        int temp1=Integer.MAX_VALUE,temp2=Integer.MAX_VALUE;        if(ops.length==0) return m*n;        for (int i = 0; i < ops.length; i++) {            if(temp1>ops[i][0]) temp1 = ops[i][0];            if(temp2>ops[i][1]) temp2 = ops[i][1];        }        return temp1*temp2;    }}